16x^2=398x-4)

Solution for 16x^2=398x-4) equation:

16x^2=398x-4)

We move all terms to the left:

16x^2-(398x-4))=0

We get rid of parentheses

16x^2-398x+4)=0

We add all the numbers together, and all the variables

16x^2-398x=0

a = 16; b = -398; c = 0;
Δ = b2-4ac
Δ = -3982-4·16·0
Δ = 158404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{158404}=398$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-398)-398}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-398)+398}{2*16}=\frac{796}{32}
=24+7/8
$